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3 years ago

Can light have inertia??

Physics

2 answers:

loris [4]3 years ago

8 0

Yes! Interim is in units of mass so yes!

Strike441 [17]3 years ago

6 0

Answer:

Yes.

Explanation:

Yes Light has inertia, since inertia is in units of mass.

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The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u

irinina [24]

Explanation :

The forces acting on hot- air balloon are:

Weight, (W)

Force due to air resistance, (F)

Upthrust force, (U)

Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.

In this case, the hot-air balloon descends vertically at constant speed.

so, $a=0$

and $F=ma=0$

so, $W = F + U$ ....................(1)

when it is ascending let the weight that it is releasing is R, so

$(W-R) + F = U$ ..........(2)

solving equation (1) and (2)

$(W-R)+F=W-F$

$R=2F$

2F is the weight of material that must be released from the balloon so that it ascends vertically at the same constant speed.

7 0

3 years ago

Read 2 more answers

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago

What is one benefit to lifelong physical activity?How can you measure your level of intensity during a physical activity?

daser333 [38]

It will lower your Chances of diseases, heart attack, and cholesterol. you can measure your intensity by making sure your body is under a safe amount of stress

5 0

3 years ago

Please need help ASAP. Would really appreciate it.

a_sh-v [17]

if i renember correctly its b

3 0

3 years ago

An object will begin moving from rest when acted upon by which forces?

denis23 [38]

Answer:

C. The gravity acceleration is in the same direction as the force of gravity, and thus toward the center of the earth.

Explanation:

[ More info.- It doesn't affect the body when it is at the rest, and when it is in motion, it acts toward the center of the Earth and it's component is known as "Centripetal force" ]

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3 years ago